Question 171: Work, Energy and Power & Circular Motion

Question

A point particle of mass $$m,$$ moves long the uniformly rough track $$PQR$$ as shown in the figure. The coefficient of friction, between the particle and the rough track equals $$\mu .$$ The particle is released, from rest from the point $$P$$ and it comes to rest at point $$R.$$ The energies, lost by the ball, over the parts, $$PQ$$ and $$QR$$, of the track, are equal to each other , and no energy is lost when particle changes direction from $$PQ$$ to $$QR$$.
The value of the coefficient of friction $$\mu $$ and the distance $$x$$ $$(=QR),$$ are, respectively close to:

Accepted Answer

Using work energy theorem for the motion of the particle,
Loss in $$P.E.=$$ Work done against friction from $$p 	o Q$$
$$ + $$ work done against friction from $$Q 	o R$$
$$mgh = \mu \left( {mg\cos 	heta } ight)PQ + \mu mg\left( {QR} ight)$$
$$h = \mu \,\cos \,	heta  	imes PQ + \mu \left( {QR} ight)$$
$$2 = \mu  	imes {{\sqrt 3 } \over 2} 	imes {2 \over {\sin \,{{30}^ \circ }}} + \mu x$$
$$2 = 2\sqrt 3 \mu  + \mu x$$  $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(i)$$
[ $$\sin \,\,{30^ \circ } = {2 \over {PQ}}$$ ]
According to the question, work done $$P 	o Q = $$ work done $$Q 	o R$$
$$ \Rightarrow $$$$\mu \left( {mg\cos 	heta } ight)PQ + \mu mg\left( {QR} ight)$$
$$	herefore$$ $$2\sqrt 3 \,\mu  = \mu x$$
$$	herefore$$ $$x \approx 3.5m$$
From $$(i)$$
$$2 = 2\sqrt 3 \mu  + 2\sqrt 3 \mu  = 4\sqrt 3 \mu $$
$$	herefore$$ $$\mu  = {2 \over {4\sqrt 3 }} = {1 \over {2 	imes 1.732}} = 0.29$$

Answer

$$0.29$$ and $$3.5$$ $$m$$

Answer

$$0.29$$ and $$6.5$$ $$m$$

Answer

$$0.2$$ and $$6.5$$ $$m$$

Answer

$$0.2$$ and $$3.5$$ $$m$$

Work, Energy and Power & Circular Motion

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